Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{p^3 - 11p^2 + 28p}{p^2 - 13p + 36} \times \dfrac{p - 9}{p^2 - 10p} $
Solution: First factor out any common factors. $n = \dfrac{p(p^2 - 11p + 28)}{p^2 - 13p + 36} \times \dfrac{p - 9}{p(p - 10)} $ Then factor the quadratic expressions. $n = \dfrac {p(p - 4)(p - 7)} {(p - 4)(p - 9)} \times \dfrac {p - 9} {p(p - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { p(p - 4)(p - 7) \times (p - 9)} { (p - 4)(p - 9) \times p(p - 10)} $ $n = \dfrac {p(p - 4)(p - 7)(p - 9)} {p(p - 4)(p - 9)(p - 10)} $ Notice that $(p - 4)$ and $(p - 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {p\cancel{(p - 4)}(p - 7)(p - 9)} {p\cancel{(p - 4)}(p - 9)(p - 10)} $ We are dividing by $p - 4$ , so $p - 4 \neq 0$ Therefore, $p \neq 4$ $n = \dfrac {p\cancel{(p - 4)}(p - 7)\cancel{(p - 9)}} {p\cancel{(p - 4)}\cancel{(p - 9)}(p - 10)} $ We are dividing by $p - 9$ , so $p - 9 \neq 0$ Therefore, $p \neq 9$ $n = \dfrac {p(p - 7)} {p(p - 10)} $ $ n = \dfrac{p - 7}{p - 10}; p \neq 4; p \neq 9 $